Find the solution of the diff eq that satisfies the given initial condition.
y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2
y' = (a+y)/tanx
dy/ (a+y) = cot x dx
ln (a + y ) = ln (sinx ) + c
I think to make this work, c will be 2a- root2/2, so my answer is
y = sinx - a + 2a - root2 /2
Is this correct?